Many collections of biomolecular fibers are modeled as semi-flexible polymer sol

Question

Many collections of biomolecular fibers are modeled as semi-flexible polymer solutions. An important length scale in such systems is the mesh size , which represents the typical spacing between polymers in solution. The following equation relates the mesh size to the fiber radius a and the fiber volume fraction :

= a /

In the adventitia of the thoracic artery of the rat, the collagen fibrils have a mean diameter of 60.0 nm with standard deviation of 10.0 nm and a mean volume fraction of 0.0812 with standard deviation of 0.0141. Calculate the mean mesh size in the adventitia of rat thoracic artery, along with its 95% confidence interval. (Note that a and are independent from each other.)

Explanation / Answer

Y = [ 1/ * sqrt(2) ] * e -(x - )2/22

for

diameter of 60.0 nm with standard deviation of 10.0 nm

radius = d/2

mean = 30, variance = 100/4 = 25

standard deviation = 5

a = [ 1/5 * sqrt(2) ] * e -(x - 30)2/50

a = e -(x - 30)2/50 / 12.55

for a mean volume fraction of 0.0812 with standard deviation of 0.0141.

=  [ 1/0.0141 * sqrt(2) ] * e -(x - 0.812)2/0.0004

= 28.3 * e -(x - 0.812)2/0.0004

= 5.32 * e -(x - 0.812)2/0.0008

now,

= a /

= (e -(x - 30)2/50 / 12.55 ) / ( 5.32 * e -(x - 0.812)2/0.0008)

= 1/66.766 * e -(x - 30)2/50* e (x - 0.812)2/0.0008

plotting the graph and comparing with standard normal form

* sqrt(2) = 66.766

= 26.63

The following distribution doesnot follow a normal distribution

it crosses x-axis at 0.663 and 0.955

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